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Post Closed as "Not suitable for this site" by abx, Carlo Beenakker, user44191, Pace Nielsen, Gerry Myerson
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u136536
u136536
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How to show that the sinc function series $\sum_{n=-\infty}^\infty\text{sinc}(xx+n)$ is equal to pi for all $x$?

We know the very well-known identity: $$\sum_{n=-\infty}^\infty\text{sinc}(n)=\pi.$$ But how to show that $$\sum_{n=-\infty}^\infty\text{sinc}(x+n)=\pi?$$ In other words, how to prove that $$\sum_{n=-\infty}^\infty\text{sinc}(x)=\pi, \qquad \forall x?$$ Any ideas?$$\sum_{n=-\infty}^\infty\text{sinc}(x+n)=\pi, \qquad \forall x?$$

How to show that the sinc function series $\sum_{n=-\infty}^\infty\text{sinc}(x)$ is equal to pi for all $x$?

We know the very well-known identity: $$\sum_{n=-\infty}^\infty\text{sinc}(n)=\pi.$$ But how to show that $$\sum_{n=-\infty}^\infty\text{sinc}(x+n)=\pi?$$ In other words, how to prove that $$\sum_{n=-\infty}^\infty\text{sinc}(x)=\pi, \qquad \forall x?$$ Any ideas?

How to show that the sinc function series $\sum_{n=-\infty}^\infty\text{sinc}(x+n)$ is equal to pi for all $x$?

We know the very well-known identity: $$\sum_{n=-\infty}^\infty\text{sinc}(n)=\pi.$$ But how to show that $$\sum_{n=-\infty}^\infty\text{sinc}(x+n)=\pi, \qquad \forall x?$$

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u136536
u136536
  • 35
  • 2

How to show that the sinc function series $\sum_{n=-\infty}^\infty\text{sinc}(x)$ is equal to pi for all $x$?

We know the very well-known identity: $$\sum_{n=-\infty}^\infty\text{sinc}(n)=\pi.$$ But how to show that $$\sum_{n=-\infty}^\infty\text{sinc}(x+n)=\pi?$$ In other words, how to prove that $$\sum_{n=-\infty}^\infty\text{sinc}(x)=\pi, \qquad \forall x?$$ Any ideas?